How to find Population mean when there are no data points ?

Not sure if this is the right place to post homework questions or not, but my math problem is this: A consumer magazine has contacted a simple random sample of 33 owners of a certain model of automobile and asked each owner how many defects had to be corrected within the first 2 months of ownership. The average number of defects was sample mean= 3.7, with a standard deviation of 1.8 defects. a. Use the t distribution to construct a 95% confidence interval for population mean =the average number of defects for the model and b. use the z distribution to construct a 95% confidence interval for population mean = the average number of defects for this model. c. given that the population standard deviation is not known, which of these two confidence intervals should be used as the interval estimate for the population mean?

So as far as I can figure I need to find the population mean first and then go back through and then do the Population standard deviation, to work out the first two parts of the problem. Unless there is another way of going about it, I'm not exactly sure though.

Ok, well I'm guessing the hypothesis is to figure out how many defects that the given sample of car owners had to get fixed in over the first 2 months that they had owned the car.

and would have to use the formula t=(sample mean(3.7)-population mean(?))/(s(?)/Square root of n(33)) So I think I would have to find the sample standard deviation, but I'm not really sure how to do that with no population mean. And from there to do the variance. I know I did something similar to this in another homework problem its just harder to pick things out in word problems like this.

Ok, well I'm guessing the hypothesis is to figure out how many defects that the given sample of car owners had to get fixed in over the first 2 months that they had owned the car.

and would have to use the formula t=(sample mean(3.7)-population mean(?))/(s(?)/Square root of n(33)) So I think I would have to find the sample standard deviation, but I'm not really sure how to do that with no population mean. And from there to do the variance. I know I did something similar to this in another homework problem its just harder to pick things out in word problems like this.

There is no hypothesis involved. If the consumer magazine wanted to see if the population mean was larger than the manufacturer's claim, then we would actually be talking about two hypotheses - a null hypothesis (e.g, mean = or < manufacturer's claim) and an alternative hypothesis (mean > manufacturer's claim).

In that situation, you, OP, would use the t-test if either the data follows a normal curve or you have a large sample (a rule of thumb is that the sample size should be >= 30), and the population standard deviation is unknown. But you're not testing a hypothesis. You've been told to construct a confidence interval.

Your book probably mentions the "t-statistic" in connection with "t-test". But a "t-statistic" has "t distribution" which has other uses. To find a "confidence interval", you don't need to know the population mean [itex] \mu [/itex]. Such a "confidence interval" has a definite length but no definite endpoints. If you look at the definition of the t-statistic, it has a term like [itex] x - \mu [/itex] in the numerator. The quantity [itex] x - \mu [/itex] can be regarded as the half length of a confidence interval with unknown center [itex] \mu [/itex]. So solve for [itex] x - \mu [/itex] as if it were a single unknown. i.e. let [itex] h = x - \mu [/itex] and find the value of h that corresponds to a probability of [itex] \frac{0.95}{2} [/itex]. You can state the confidence interval as 2h or [itex] [\mu - h, \mu+h] [/itex], depending on the conventions that your book uses. You don't have to say what [itex] \mu [/itex] is.